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%I #44 Sep 08 2022 08:45:59
%S 0,10,48,114,208,330,480,658,864,1098,1360,1650,1968,2314,2688,3090,
%T 3520,3978,4464,4978,5520,6090,6688,7314,7968,8650,9360,10098,10864,
%U 11658,12480,13330,14208,15114,16048,17010,18000,19018,20064,21138,22240
%N a(n) = 14*n^2 - 4*n.
%C Sequence found by reading the line from 0, in the direction 0, 10, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-axis of the square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].
%H Vincenzo Librandi, <a href="/A195023/b195023.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1).
%F a(n) = 2*A135703(n). - _Bruno Berselli_, Oct 13 2011
%F From _Colin Barker_, Apr 09 2012: (Start)
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F G.f.: 2*x*(5+9*x)/(1-x)^3. (End)
%t Table[14n^2-4n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,10,48},50] (* _Harvey P. Dale_, Sep 05 2012 *)
%o (Magma) [14*n^2 - 4*n: n in [0..50]]; // _Vincenzo Librandi_, Oct 14 2011
%o (PARI) a(n)=14*n^2-4*n \\ _Charles R Greathouse IV_, Apr 10 2012
%Y Cf. A144555, A152760, A195019, A195020, A195024, A195320, A185019, A193053, A198017.
%K nonn,easy
%O 0,2
%A _Omar E. Pol_, Oct 13 2011
%E Corrected by _Vincenzo Librandi_, Oct 14 2011
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